Math 215 Exam #1 Practice Problem Solutions

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1 Math 5 Exam # Practice Problem Solutions For each of the following statements, say whether it is true or false If the statement is true, prove it If false, give a counterexample (a) If A is a matrix such that A(Ax) = for all x R, then A is the zero matrix Answer: False If A(Ax) = for all x, then the column space of A and the nullspace of A must be the same space In particular, consider the matrix Then, for any x = and x A = R x, we have that Ax = A(Ax) = x = x x x = Hence, A(Ax) = for all x, but A, so A gives a counterexample to the statement (b) A system of equations in 4 unknowns can never have a unique solution Answer: True We can realize such an system of equations as a single matrix equation Ax = b, where A is a 4 matrix Hence, rank(a), so the dimension of the nullspace of A is at least : dim nul(a) = 4 rank(a) 4 = Hence, there must be at least one free variable in the system, meaning that, if the system is solvable at all, it must have an infinite number of solutions (c) If V is a vector space and S is a finite set of vectors in V, then some subset of S forms a basis for V Answer: { } False Let V = R, which is clearly a vector space, and let S be the singleton set The single element of S does not span R, so no subset of S can be a basis for R Hence, this provides a counterexample to the statement (d) Suppose A is an m n matrix such that Ax = b can be solved for any choice of b R m Then the columns of A form a basis for R m Answer: False Consider the matrix A = Then A is already in reduced echelon form and clearly has pivots, so rank(a) = This implies that dim col(a) =, so the column space of A consists of all of R Thus, the equation Ax = b can be solved for any b R (since any b is in col(a)) However, the columns of A are clearly not linearly independent (no set containing the zero vector can be linearly independent), so they cannot form a basis for R A related but true statement would be the following: Suppose A is an m n matrix such that Ax = b can be solved for any choice of b R m Then some subset of the columns of A forms a basis for R m

2 (e) Given equations in 4 unknowns, each describes a hyperplane in R 4 If the system of those equations is consistent, then the intersection of the hyperplanes contains a line Answer: True This is really just a restatement of (b) Translating the system of equations into a matrix equation Ax = b, the nullspace of A must be at least one-dimensional, so the solution-space must be at least one-dimensional Since the solution space of the matrix equation corresponds to the intersection of the hyperplanes, that intersection must be at least one-dimensional, meaning it must contain a line (f) If A is a symmetric matrix (ie A = A T ), then A is invertible Answer: False Consider the symmetric matrix A = Then A only has rank, meaning that A cannot be invertible, so this gives a counterexample to the statement (g) If m < n and A is an m n matrix such that Ax = b has a solution for all b R m, then there exists z R m such that Ax = z has infinitely many solutions Answer: True The fact that Ax = b has a solution for all b R m means that the column space of A is equal to all of R m Hence, Since rank(a) = dim col(a) = m dim nul(a) = n rank(a) = n m and since m < n, we have that the nullspace of A has some positive dimension Since the nullspace of A consists precisely of those x R n such that Ax =, this equation has infinitely many solutions Thus, letting z =, we see that the statement is true (h) The set of polynomials of degree 5 forms a vector space Answer: True You should check that the set of polynomials of degree 5 satisfies all the rules for being a vector space The important facts are this space is closed under addition and scalar multiplication For each of the following, determine whether the given subset is a subspace of the given vector space Explain your answer (a) Vector Space: R 4 Subset: The vectors of the form Answer: Yes, this is a subspace If we take two vectors in the subset, say a b d a b and a b, then their sum a b d + a b d = is also in the subset, so this set is closed under addition a + a b + b d + d d d

3 Moreover, if c R, then c a b d = is in the set, so this set is closed under scalar multiplication Thus, the set is closed under both addition and scalar multiplication, and so is a subspace (b) Vector Space: R Subset: The solutions to the equation x 5y = Answer: No, this is not a subspace To see why, I ll show that it is not closed under addition The vectors and are both in the set, since the pairs (/, ) and (, /5) both solve the equation x 5y =, but + = 5 5 is not in the set, since ca cb cd 5 (/) 5( /5) = + = Therefore, the set is not closed under addition, and so is not a subspace (c) Vector Space: R n Subset: All x R n such that Ax = x where A is a given n n matrix Answer: Yes, this is a subspace To prove it, suppose x and x are in this set, meaning that Ax = x and Ax = x (such vectors are called eigenvectors of A; we ll learn more about them later) Then A(x + x ) = Ax + Ax = x + x = (x + x ), meaning that x + x is in this set as well Moreover, for any c R, A(cx ) = c(ax ) = c(x ) = (cx ), so cx is in the set as well Therefore, this set is closed under addition and scalar multiplication, so it is indeed a subspace (d) Vector Space: R Subset: The intersection of P and P, where P and P are planes through the origin Answer: Yes, this is a subspace The proof is essentially the same as you gave for Problem (c) from HW 4 (e) Vector Space: All polynomials Subset: The quadratic (ie degree ) polynomials Answer: No, this is not a subspace To see that it is not closed under addition, notice that if f(t) = t and g(t) = t, then f and g are both in the set of quadratic polynomials, but, since the sum f + g is not a quadratic polynomial (f + g)(t) = f(t) + g(t) = t + ( t ) =,

4 (f) Vector Space: All real-valued functions Subset: Functions of the form f(t) = a cos t + b sin t + c for a, b, c R Answer: Yes, this is a subspace If a, a, b, b, c, c R and I define and f(t) = a cos t + b sin t + c g(t) = a cos t + b sin t + c, then f and g are in the given subset The sum has the form f(t)+g(t) = (a cos t+b sin t+c )+(a cos t+b sin t+c ) = (a +a ) cos t+(b +b ) sin t+(c +c ), so f + g is also in the subset, which is, therefore, closed under addition Also, if r R, then rf(t) = r(a cos t + b sin t + c ) = (ra ) cos t + (rb ) sin t + (rc ), so rf is in the subset, which is, therefore, closed under scalar multiplication Hence, we can conclude that this subset is actually a subspace Consider the matrix A = a a (a) Under what conditions on a is A invertible? Answer: The matrix A is invertible if and only if it has rank To see what the rank is, we do elimination The first step is to subtract a times row from row, yielding a a Then this has a second pivot if and only if a, meaning that a, or a ± Thus, A is invertible so long as a is neither nor (b) Choose a non-zero value of a that makes A invertible and determine A Answer: Choose a = Recall that we can find the inverse of A by converting the left side of the following augmented matrix to the identity: Subtract twice row from row : Scale the second row by Therefore, and also subtract twice the result from row : A = 4

5 (c) For each value of a that makes A non-invertible, determine the dimension of the nullspace of A Answer: When a =, the matrix A =, which, after subtracting row from row, reduces to Hence, A has rank, so the nullspace has dimension 4 Consider the system of equations dim nul(a) = rank(a) = = x + x + x x 4 = b x + x + x 5x 4 = b x + 4x + x 8x 4 = b (a) Find all solutions when the above system is homogeneous (ie b = b = b = ) Find a basis for the space of solutions to the homogeneous system Answer: Convert the system into the augmented matrix Now do elimination to get the reduced echelon form First, subtract row from row and subtract twice row from row : Now, subtract row from both row and row : Then this system is consistent provided that x = x + x 4 x = x 4 Hence, the solutions to the homogeneous equation are those vectors of the form x + x 4 for x, x 4 R Then a basis for the space of solutions to the homogeneous system (ie nullspace of the corresponding matrix) is, 5

6 (b) Let S be the set of vectors b = b b such that the system can be solved What is the dimension b of S? Answer: Letting A be the matrix of the system, we know that the set of vectors b for which the system can be solved is the column space of A Since A is 4, we know that rank(a) + dim nul(a) = 4 Since, from part (a), we know that the dimension of the nullspace is, this implies that the column space of A is two-dimensional (c) It s easy to check that the vector v = is a solution to the system that arises when b =, b = 5, and b = 8 Find all the solutions to this system Answer: All solutions x to the system Ax = b take the form x = x p +x h, where x p is a particular solution and x h is the homogeneous solution to the corresponding homogeneous problem Thus, we can let x p = v, which we re told solves the system and we see that, using part (a), the general solution is where x, x 4 R + x + x 4, 6

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